Existing of such automorphism subgroup

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Let $G$ be abelian group and $Aut(G)$ be the automorphism group of $G$ ,

I am looking for a nontrivial subgroup $H$ of $Aut(G)$ such that $gcd(|H|,|G|)=1$.

Does such subgroups always exit ?

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Since $\operatorname{Aut}(A \times B) \cong \operatorname{Aut}(A) \times \operatorname{Aut}(B)$ when $(|A|, |B|) = 1$, it is enough to consider the case where $G$ has prime power order $p^\alpha$. In this case the question is whether $\operatorname{Aut}(G)$ has order divisible by a prime $q \neq p$.

This is not always true, for example $G = C_2$ has a trivial automorphism group and $G = C_2 \times C_4$ has automorphism group isomorphic to $D_8$.

Actually this sort of thing can only happen for $p = 2$, and we can classify all such groups. They are all of the form

$$C_{2^{e_1}} \times C_{2^{e_2}} \times \cdots \times C_{2^{e_n}}$$

where $n \geq 0$ and $1 \leq e_1 < e_2 < \cdots < e_n$ are integers.

This follows from the formula for $|\operatorname{Aut}(G)|$ given in Theorem 4.1 of this article.